a special Olympics event has 16 more boys than girls competing. the total number of participants is 1000. how many boys competed and how many girls competed?

Question

Algebra word problem

Answer 1

Hi Samantha,

 

Let's begin by assigning letters to our two unknowns:

    b = number of boys

    g = number of girls

 

Now let's come up with two equations which we can use to solve for our two unknowns:

    Equation #1:  b = g + 16 (# of boys vs. # of girls)

    Equation #2:  b + g = 1000 (total # of kids)

 

Now let's plug in our expression for # of boys (b) in terms of # of girls (g) from Equation #1 into Equation #2 and solve for # of girls (g):

    g + 16 + g = 1000

    2g = 1000 - 16

    2g = 984

      g = 984 / 2

      g = 492 (# of girls)

 

Now let's plug our value for # of girls (492) into Equation #1 and solve for # of boys (b):

    b = 492 + 16

    b = 508 (# of boys)

 

Finally, we can plug both our values for # of boys (508) and # of girls (492) into Equation #2 and see if both sides of the equation are equal:

    508 + 492 = 1000

    1000 = 1000 (both sides equal)

 

Since both sides of the equation are equal, we have successfully confirmed that both our values are correct.

 

Thanks for submitting this problem and glad to have been of help.

 

God bless, Jordan. 

 

Answer 2

g is the number of girls

g + 16 is the number of boys

 

g + g + 16 = 1000

2g = 984

g = 492

 

492 girls competed along with 508 boys.

 

Check

492 + 508

1000

Answer 3

Let:
x = number of girls
y = number of boys

Information given in problem:
number of boys is 16 more than girls --> y = x + 16
total number of participants is 1000   --> y + x = 1000

Using substitution:
y + x = 1000
x + 16 + x = 1000 (b/c y = x + 16)
2x + 16 = 1000
     - 16       -16
2x         = 984

divide each side by 2 to get x = 492.

There fore 492 girls and 508 boys competed.